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2 years ago

What is the acceleration of a 160kg object if a force of 150N is applied to it?

Physics

1 answer:

Nadusha1986 [10]2 years ago

3 0

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

m is the mass

f is the force

From the question we have

$a = \frac{150}{160} = \frac{15}{16} \\ = 0.9375$

We have the final answer as

Hope this helps you

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A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.

fgiga [73]

The vertical velocity of the projectile upon returning to its original is 17. 74 m/s

<h3>How to determine the vertical velocity</h3>

Using the formula:

Vertical velocity component , Vy = V * sin(α)

Where

V = initial velocity = 36. 6 m/s

α = angle of projectile = 29°

Substitute into the formula

Vy = 36. 6 * sin ( 29°)

Vy = 36. 6 * 0. 4848

Vy = 17. 74 m/s

Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s

Learn more about vertical velocity here:

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1 year ago

Which planet has the GREATEST attraction to the sun?

AveGali [126]

I think its Mercury because it's the closest to the sun.

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3 years ago

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Rom4ik [11]

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3 years ago

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A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre

Alisiya [41]

Answer:

0.546 $\hat k$

Explanation:

From the given information:

The force on a given current-carrying conductor is:

$F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})$

where the length usually in negative (x) direction can be computed as

$\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i$

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

$\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})$

$F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)$

$F = I \int^3_1 - 9.0x^2 \ dx \hat k$

$F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k$

$F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k$

where;

current I = 7.0 A

$F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k$

$F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k$

F = 546 × 10⁻³ T/mT $\hat k$

F = 0.546 $\hat k$

4 0

3 years ago

A long solenoid of radius 3 cm has 1100 turns per meter. If the solenoid carries a current of 1.5 A, then calculate the magnetic

Arada [10]

Answer:

The magnetic field at the center of the solenoid is 2.1 × 10⁻³ T

Explanation:

The magnetic field B at the center of the solenoid is given by

B = μ₀ni where μ₀ = permeability of free space = 4π × 10⁻⁷H/m, n = number of turns per unit length of the solenoid = 1100 turns per meter and i = current in the solenoid = 1.5 A.

So B = μ₀ni

= 4π × 10⁻⁷H/m × 1100 × 1.5 A

= 4π × 10⁻⁷H/m × 1650 A-turns/m

= 20734.5 × 10⁻⁷T

= 2.07345 × 10⁻³ T

≅ 2.1 × 10⁻³ T

So the magnetic field at the center of the solenoid is 2.1 × 10⁻³ T

3 0

3 years ago